Level-2 (Application/Understanding-based) questions from the Class 12 Chemistry Chapter – Solutions

 

5 more Level-2 (Application/Understanding-based) questions from the Class 12 Chemistry Chapter – Solutions, along with detailed answers:

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Q1. A 0.1 molal solution of a non-electrolyte in water freezes at –0.186°C. Calculate the molal depression constant (Kf) of water. (Freezing point of pure water = 0°C)

Answer:

$$\Delta T_f = T_f^0 - T_f = 0 - (-0.186) = 0.186^\circ C$$ $$\Delta T_f = K_f \cdot m \Rightarrow K_f = \frac{\Delta T_f}{m} = \frac{0.186}{0.1} = 1.86\, \text{K·kg/mol}$$

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Q2. Explain why aquatic species are more comfortable in cold water than in warm water.

Answer:
According to Henry's Law, the solubility of gases in liquids decreases with an increase in temperature.
Thus, in cold water, more oxygen dissolves, making it more suitable for aquatic organisms.
Warm water holds less dissolved oxygen, which may lead to suffocation of aquatic life.

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Q3. A solution is prepared by dissolving 5 g of a non-volatile solute in 95 g of water. The vapor pressure of pure water at 298 K is 23.8 mmHg. The vapor pressure of the solution is found to be 23.4 mmHg. Calculate the molar mass of the solute.

Answer:

$$\text{Relative lowering of vapor pressure} = \frac{P^0 - P}{P^0} = \frac{23.8 - 23.4}{23.8} = \frac{0.4}{23.8} \approx 0.0168$$

Using:

$$\frac{n_{\text{solute}}}{n_{\text{solvent}} + n_{\text{solute}}} \approx \frac{n_{\text{solute}}}{n_{\text{solvent}}} = 0.0168$$ $$n_{\text{solvent}} = \frac{95}{18} \approx 5.28 \, \text{mol}$$ $$\frac{w}{M} = 0.0168 \times 5.28 \Rightarrow \frac{5}{M} = 0.0887 \Rightarrow M \approx \frac{5}{0.0887} \approx 56.37 \, \text{g/mol}$$

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Q4. What is the effect of adding NaCl to water on its boiling point and freezing point? Explain with reason.

Answer:

• Boiling point increases – Adding NaCl increases the number of solute particles, leading to elevation in boiling point (a colligative property).

• Freezing point decreases – NaCl causes depression in freezing point due to disruption of ice crystal formation.

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Q5. Calculate the mass of NaCl (i = 2) to be dissolved in 1000 g of water to decrease its freezing point by 3.72°C. (Kf for water = 1.86 K·kg/mol)

Answer:

$$\Delta T_f = i \cdot K_f \cdot m \Rightarrow 3.72 = 2 \cdot 1.86 \cdot m \Rightarrow m = \frac{3.72}{3.72} = 1\, \text{mol/kg}$$

Mass of NaCl for 1 mol = 58.5 g
∴ Required mass = 58.5 g